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The Weyl integral formula states that if $f$ is a class function on $U(n),$ $T$ is the torus of diagonal matrices in $U(n)$, and $dU(n)$ and $dT$ are the standard Haar measures on $U(n)$ and $T,$ then:

$$\int_{U(n)} f(g)\; dU(n) = \frac{1}{n!} \int_T f(z_1, \ldots, z_n) \left \lvert \prod_{i<j} (z_i-z_j)\right \rvert^2 \;dT.$$

The Weyl character formula follows quite quickly from the WIF, by just calculating which virtual characters have inner product 1 with themselves.

It is clear that there should be some formula along the lines of the WIF, where the norm of the Weyl denominator $|\Delta|^2$ is replaced by the formula for the volume of each conjugacy class. The exact fromula can be proved by a not terribly difficult direct calculation (e.g. these notes). But Dylan Thurston pointed out to me last week that there's a nice non-computational heuristic outline where you just find out what properties $|\Delta|^2$ has to have, and then guess that it should be the simplest formula satisfying those properties.

My question is whether that heuristic can be made rigorous. I have in mind something like the following outline. We'll let $V(z_1, \ldots, z_n)$ denote the "volume of the conjugacy class" function that we're trying to show is $|\Delta|^2$:

  1. By "algebraic geometry" $V$ should be a polynomial in the $z_i$ and $\bar{z_i}$.
  2. Since the dimension of the conjugacy class drops when $z_i = z_j$, the polynomial $V$ should be divisible by $\Delta$. Since the size of the conjugacy class is invariant under conjugation it should also be divisible by $\bar{\Delta}$ and hence by $\Delta \bar{\Delta}$.
  3. By "dimensional analysis" $V$ should be homogenous of degree the dimension of the conjugacy class which is $\dim U(n) - \dim T = n^2-n = \deg |\Delta|^2$. Hence up to scalar $V$ must be $\Delta \bar{\Delta}$.
  4. The scalar must be $1/n!$ by using the fact that the trivial representation is irreducible and computing its inner product with itself using the WIF.

Is there a rigorous proof of the WIF along these lines? Clearly step 3 is very sketchy, and probably totally bogus.

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    $\begingroup$ What's going to make this difficult, I think, is that for any other compact group $K$ the conjugacy classes are not complex algebraic varieties. For example, $Spin(5)$ has $Spin(5)/Spin(4) = S^4$ as a conjugacy class. (Only in type $A$ do all the conjugacy classes have $K$-invariant complex structures.) $\endgroup$
    – Allen Knutson
    Oct 8, 2015 at 18:37
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    $\begingroup$ I'm not totally sure I understand why that would be relevant. After all the formula for V isn't complex algebraic, it's only real algebraic, right? $\endgroup$
    – Noah Snyder
    Oct 8, 2015 at 19:34
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    $\begingroup$ Part (1) seems sketchier to me than part (3). I can make part (3) precise by considering the volume of $\{ U \mathrm{diag}(z_1, z_2, \ldots, z_n) U^{-1} \}$ for arbitrary complex numbers $(z_1, \ldots, z_n)$, not just roots of unity. Once I do that, the formula is literally homogenous of degree $n^2-n$ when I rescale the $z_i$, because I literally rescale the orbit. $\endgroup$
    – David E Speyer
    Oct 8, 2015 at 20:34
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    $\begingroup$ Why polynomial rather than rational? Is that what the last couple comments are about? $\endgroup$
    – Ben Wieland
    Oct 11, 2015 at 20:54
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    $\begingroup$ @BenWieland: Good question... I think the usual Cramer's rule formula for inverses tells you that you that the only denominators that can come up are powers of the determinant. In the determinant 1 case that meant they were polynomials, but one does need to be a little careful now with David's generalized version to make sure it really is a polynomial. $\endgroup$
    – Noah Snyder
    Oct 13, 2015 at 3:25

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